Difficult
The temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$
- A$1100$
- B$1200$
- C$1303$
- D$1405$
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Similar Questions
Activation energy of a chemical reaction can be determined by
MediumAIPMT 1998
View SolutionThe following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$
For a given reaction,the overall enthalpy change is $+100 \ kJ/mol$ and the activation energy of the reverse reaction is $+200 \ kJ/mol$. The activation energy for the forward reaction would be......$kJ/mol$.
Medium
View SolutionWhat is Activation Energy $(E_a)$? Explain the reaction profile graph for activation energy and discuss the probability of effective collisions.
Difficult
View SolutionThe graph for the process $R \rightarrow P$ is given below. Determine the activation energy $(E_a)$ for the forward reaction if the energy of the reactant is $50 \ kJ$ and the energy of the transition state is $170 \ kJ$. (in $kJ$)
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